Smash Products of Multiplier Left Hopf Algebras

Abstract

Firstly, we define and study the notions of a smash product for actions of multiplier left Hopf algebras on algebras and of an integral on such smash products. Then we construct an analogue of Radford’s biproduct in the framework of multiplier left Hopf algebras under assumption of a multiplier left Hopf algebra having an anti-bialgebra homomorphic left antipode. Finally, we study a duality theorem for smash products of a left Hopf algebra of dimension n which is a special multiplier left Hopf algebra.

1. Introduction

In Hopf algebra theory, on the one hand, smash products in [1] play a very important role to establish a new Hopf algebra from two given Hopf algebras. This approach was extended to multiplier Hopf algebras in order to get new multiplier Hopf algebras (see [2]). On the other hand, The Blattner–Montgomery (BM) duality theorem is a corner stone of Hopf algebra theory, originating from operator algebra and evolving into a unifying principle for quantum groups and category theory. The duality principle began with Takesaki’s theorem for locally compact groups in 1973 [3,4,5]:

$$(N⋊G)⋊\widehat{G}\cong N\overline{\otimes }\mathcal{B}\left(L^2\left(G\right)\right).$$

For group algebras, Cohen and Montgomery proved, in 1976, the algebraic prototype for finite groups:

$$(R\#\mathbf{k}G)\#{\left(\mathbf{k}G\right)}^{*}\cong R\otimes M_n\left(\mathbf{k}\right).$$

Blattner and Montgomery (BM) established in 1985 the general Hopf algebra version (see [6]):

$$(R\#H)\#H^{\circ }\cong R\otimes (H\#H^{\circ }).$$

Particularly, in the case of a finite dimensional (f.d.) of dimension n for H, the above result becomes $(A\#H)\#H^{*}\cong M_n\left(A\right)$ (BM duality). This duality theorem was still true within the framework of Hopf quasimodule algebras [7]. After that, there are the following generalizations (1980s–resent):

• 1985: Radford Biproducts linked duality to Yetter–Drinfeld modules.
• 1990: Majid’s Double Cross Products unified the Drinfeld Double $D\left(H\right)$.
• 1990s–2000s: Extended to braided, weak, quasi-Hopf algebras.
• 2010s–Present: Hopf quasigroups, categorified into fusion categories and higher algebra.

It is worth mentioning that, in our previous article [7], assuming H is a Hopf quasigroup [8], i.e., an associative algebra (generally not bialgebra) with unit 1 armed with counital algebra homomorphisms $\Delta \in Ho{m}_{\mathbf{k}}(H,H\otimes H)$ and $\epsilon \in Ho{m}_{\mathbf{k}}$ and a $\mathbf{k}$-map $S:H⟶H$ so that $\sum S\left(h^{\prime }\right){\left(h^{″}\right)}^{\prime }\otimes {\left(h^{″}\right)}^{″}=1\otimes h=\sum h^{\prime }S\left({\left(h^{″}\right)}^{\prime }\right)\otimes {\left(h^{″}\right)}^{″}$ and $\sum {\left(h^{\prime }\right)}^{\prime }\otimes S\left({\left(h^{\prime }\right)}^{″}\right)h^{″}=h\otimes 1=\sum {\left(h^{″}\right)}^{\prime }\otimes {\left(h^{″}\right)}^{″}S\left(h^{″}\right)$, for $h\in H$ (here we write $\Delta \left(h\right)=\sum h^{\prime }\otimes h^{″}$), then the authors in [7] proved that the BM duality theorem holds.

Left Hopf algebras (see [9]) are very different from Hopf quasigroups (see [7]); they are a class of bialgebras with left antipodes, from which many left quantum groups can be constructed. Obviously, the dual of a finite-dimensional left quantum group is also a left quantum group, and the Pontryagin duality theorem holds. However, for infinite-dimensional left quantum groups, its linear dual is not a left quantum group. In this case, to establish the Pontryagin duality theorem, in the literature [10], the authors of this paper introduced the notions of multiplier left Hopf algebras and studied their properties as a promotion of a left Hopf algebra (see [9]). Obviously, multiplier left Hopf algebras are also different from Hopf quasigroups and Hopf algebras.

Then, how do we define the smash product of multiplier left Hopf algebras and when does the smash product of two multiplier left Hopf algebras have the structure of a multiplier left Hopf algebra, so that we can construct more examples of multiplier left Hopf algebras? We also ask: does the above duality theorem hold for left Hopf algebras?

These questions become a motivation for writing this article. The article is arranged as follows.

The first section mainly reviews some necessary materials about left multiplier Hopf algebras. The main purpose of the second section is to study the smash product for multiplier left Hopf algebras and construct an analogue of Radford’s biproduct in the setting of multiplier left Hopf algebras under assumption of a multiplier left Hopf algebra having an anti-bialgebra homomorphic left antipode (see Theorem 1). From this, we can construct non-trivial examples of multiplier left Hopf algebras. Furthermore, we also consider integrals on the smash product (see Proposition 2).

Finally, the main research and proof of the third and fourth sections are BM’s theorem under the framework of f.d. left Hopf algebra (Theorem 2, Corollary 3).

In this article, we always work over a fixed field $\mathbf{k}$. For a $\mathbf{k}$-map $\Delta \in Ho{m}_{\mathbf{k}}(C,C\otimes C)$, we write (cf. [11]): $\Delta \left(c\right)=\sum c^{\prime }\otimes c^{″}$, so we have $(id\otimes \Delta )\Delta \left(c\right)=(\Delta \otimes id)\Delta \left(c\right)=\sum c^{\prime }\otimes c^{″}\otimes c^{‴}$ for $c\in C$, and so on.

2. Preliminaries

In this section, for convenience, we collect some concepts and conclusions for multiplier (left) Hopf algebras and left Hopf algebras (see [9,10]).

2.1. Multiplier (Left) Hopf Algebras

Assume that $(A,m)$ is a nondegenerate algebra, with a product m, but A cannot be an identity element. Let $\Delta :A⟶M(A\otimes A)$ be a coproduct on $(A,m)$ in the sense of ([12], Definition 1.1). We define the following four maps $G^{lr},G^{rl},G^{rr},G^{ll}\in Hom(A\otimes A,M(A\otimes A))$ by [13,14,15,16]

$$\begin{array}{ccc}& & G^{lr}(a\otimes b)=\Delta \left(a\right)(1\otimes b)\text{and}{G}^{rl}(a\otimes b)=(a\otimes 1)\Delta \left(b\right),\hfill \\ & & G^{rr}(a\otimes b)=(1\otimes b)\Delta \left(a\right)\text{and}{G}^{ll}(a\otimes b)=\Delta \left(b\right)(a\otimes 1).\hfill \end{array}$$

The above maps are said to be regular if they have a range in $A\otimes A$. The $\Delta$ is coassociative ([13]) if

$$(G^{rl}\otimes id)(id\otimes G^{lr})=(id\otimes G^{lr})(G^{rl}\otimes id).$$

An element $\epsilon \in Hom(A,\mathbf{k})$ is said to be a left counit if $(\epsilon \otimes \iota )G^{lr}(x\otimes y)=xy$ for $x,y\in A$. We say that A is a multiplier Hopf algebra if (i) $G^{lr}(A\otimes A)$ is in $A\otimes A$ and $G^{rl}(A\otimes A)$ is in $A\otimes A$, and $G^{lr}$ and $G^{rl}$ are bijective (see, [13]).

Definition 1

([10], Definition 3.3 (i)). $(A,m,\Delta )$ is called a multiplier left Hopf algebra (MLHA) if the following axioms are satisfied:

(i) A has a right counit;

(ii) $G^{lr}(A\otimes A)\subseteq A\otimes A$ and the maps $G^{lr}$ are injective maps of $A\otimes A$ to itself;

(iii) $G^{rl}(A\otimes A)\subseteq A\otimes A$ and $G^{rl}$ is surjective.

We say that A is regular if $A^{cop}$, where ${\Delta }^{cop}$ is obtained from Δ by flipping the factors in the tensor product, which is again a coproduct such that $(A,{\Delta }^{cop})$ is also a multiplier left Hopf algebra.

2.2. Left Hopf Algebras

Let $H:=(H,\Delta ,1_H,\epsilon )$ be a bialgebra, where $\Delta$ is a comultiplication, $1_H$ is unit, and $\epsilon$ is a counit. Then H is a left Hopf algebra (see [9]) if there exists an element $S\in End\left(H\right)$ satisfying: $\sum S\left(h^{\prime }\right)h^{″}=\epsilon \left(h\right)1_H$ and $\sum \epsilon \left(h^{\prime }\right)h^{″}=\sum h^{\prime }\epsilon \left(h^{″}\right)=h$ with $h\in H$.

Remark 1.

(1) The left Hopf algebras (not Hopf algebras) were constructed in [9] which are the first examples and free left Hopf algebras on the comatrices coalgebra of $n\times n$ for $n\ge 2$. There are some special left antipodes which are an algebra, and also a coalgebra anti-homomorphism.

(2) If $(H,S)$ is a left Hopf algebra and S is not a right antipode, then H will have an infinite number of left antipodes.

(3) For a left (right) Hopf algebra, generally, the antipode is not necessary to be a bialgebra anti-homomorphism.

3. Smash Products for Multiplier Left Hopf Algebras

In this section, A generally indicates a multiplier left Hopf algebra with an antipode S which is a bialgebra anti-homomorphism. We mainly show an analogue of Radford’s biproduct (see, [17]).

Definition 2.

Given a nondegenerate bialgebra B without unit and with counit ${\epsilon }_B$. Suppose that A acts on B, denoted as $A\succ B$. Then

(1) B is a left A-module algebra if also

$$\begin{array}{ccc}& & a\succ \left(xy\right)=\sum (a^{\prime }\succ x)(a^{″}\succ y)\hfill \end{array}$$

(1)

$$\begin{array}{ccc}& & a\succ 1_{M\left(B\right)}={\epsilon }_A\left(a\right)1_{M\left(B\right)}\hfill \end{array}$$

(2)

for $a\in A$ and $x,y\in B$. Remark that a left action of A on B can be lifted to be on $M\left(A\right)$ (see [18]), and similarly for a right version.

(2) B is called a left A-module coalgebra if also

$$\begin{array}{ccc}& & {\Delta }_B(a\succ x)=\sum (a^{\prime }\succ x^{\prime })(a^{″}\succ x^{″})\hfill \end{array}$$

(3)

$$\begin{array}{ccc}& & {\epsilon }_B(a\succ x)={\epsilon }_B\left(a\right){\epsilon }_B\left(x\right)\hfill \end{array}$$

(4)

for any $h\in H$ and $a\in A$.

(3) Let B be an MLHA. Then one says that B is an admissible A-module coalgebra if B is a left A-module algebra, and at same time a left A-module coalgebra.

Example 1.

(1) Given $B:=A$ as above.

(i) Define $a{d}^l:A\otimes B⟶B$ by $a\succ x=\sum a^{\prime }xS\left(a^{″}\right)$ with $a\in A$ and $x\in B$. In this formula, $a^{\prime }$ is covered by x (through multiplication). Then B itself becomes a left A-module algebra under ≻. However it is not a left H-module algebra in the ordinary sense in [1] since $a\succ 1\ne \epsilon \left(a\right)1$ when A is a left Hopf algebra.

(ii) Naturally, B is a left A-module coalgebra through the multiplication.

(2) Let $dim\left(A\right)<\infty$. Then $A^{*}$, the linear dual of A, is also a left Hopf algebra with antipode $S^{*}$. In fact, we have the bilinear form $\mathsf{\Omega }\in Hom(H^{*}\otimes H,\mathbf{k})$ defined via $\mathsf{\Omega }(a^{*},a)=a^{*}\left(a\right)$ with $a^{*}\in A^{*}$ and $a\in A$.

As for $a^{*}\in A^{*}$ and $a\in A$, one obtains $\mathsf{\Omega }(S^{*}\left(a^{*}\right),a)=\mathsf{\Omega }(a^{*},S\left(a\right))$ and $a^{*}\succ a$ is described by

$$a^{*}\succ a=\sum \mathsf{\Omega }(a^{*},a^{″})a^{\prime }.$$

The A becomes a left $A^{*}$-module algebra. In fact, for $ab\in A$ and $a^{*}\in A^{*}$,

$$\begin{array}{ccc}\hfill a^{*}\succ \left(ab\right)& =& \sum \mathsf{\Omega }(a^{*},a^{″}{b}^{″})a^{\prime }{b}^{\prime }=\sum \mathsf{\Omega }(a^{*\prime },a^{″})\mathsf{\Omega }({a^{*}}^{″},b^{″})a^{\prime }{b}^{\prime }\hfill \\ & & =\sum ({a^{*}}^{\prime }\succ a)({a^{*}}^{″}\succ b),\hfill \end{array}$$

and $a^{*}\succ 1=\mathsf{\Omega }(a^{*},1)1={\epsilon }^{*}\left(a^{*}\right)1$.

For an A-module algebra B, we can define a smash product $B⊡A:=B\otimes A$ as vector spaces with a multiplication described as follows:

$$(x⊡a)(y⊡b)=\sum a(a^{\prime }\succ y)⊡a^{″}b$$

(5)

for $a,b\in A$ and $x,y\in B$. In this formula, $a^{\prime }$ is covered by y (through the action) and $a^{″}$ is covered by y (through multiplication).

Proposition 1.

(1) The $B⊡A$ is an associative non-degenerate algebra.

(2) The $B⊡A$ is isomorphic with the algebra, generated by A and B, subject to the following formula:

$$ax=\sum (a^{\prime }\succ x)a^{″}$$

for any $a\in A$ and $x\in B$.

Proof. 

(1) Let $a,b,c\in A$ and $x,y,z,\in B$. One does a calculation as follows:

$$\begin{array}{ccc}& & \left[(x⊡a)(y⊡b)\right](z⊡c)=\sum [x(a^{\prime }\succ y)⊡a^{″}b](z⊡c)\hfill \\ & =& \sum a(a^{\prime }\succ y)[\left(a^{″}{l}^{\prime }\right)\succ c]⊡a^{‴}{l}^{″}c\hfill \\ & =& \sum a(a^{\prime }\succ \left(y(b^{\prime }\succ z)\right)⊡a^{″}{b}^{″}c\hfill \\ & =& \sum (x⊡a)[y(b^{\prime }\succ z)⊡b^{″}c]=(x⊡a)\left[(y⊡b)(z⊡c)\right].\hfill \end{array}$$

For the non-degenerateness, let $\sum p_{\left(i\right)}\otimes q_{\left(i\right)}\in B⊡A$, and assume that

$$(\sum _i{p}_{\left(i\right)}⊡q_{\left(i\right)})(x⊡a)=0$$

for all $a\in A,x\in B$. We have immediately ${\sum }_i{p}_{\left(i\right)}(q_i^{\prime }\succ x)⊡q_i^{″}a=0$. Following the reference [18], we can get that $\sum p_{\left(i\right)}\otimes q_{\left(i\right)}=0$, i.e., the product on $A⊡H$ is non-degenerate.

(2) Obvious. □

Remark 2.

If A is a multiplier left Hopf ∗-algebra and if B is a ∗-algebra, we require the extra condition that ${(a\succ x)}^{*}=S{\left(a\right)}^{*}\succ x^{*}$ with $a\in A$ and $x\in B$. One can show that with this extra assumption, the $B⊡A$ can constitute ∗-algebra simply by letting ${\left(ax\right)}^{*}=x^{*}{a}^{*}$ for $a\in A$ and $x\in B$.

Given two MLHAs A and B which act each other, under reasonable conditions, we will give an approach to make a new MLHA as follows.

Theorem 1.

Let A and B be two MLHAs with respective antipodes $S_A$ and $S_B$. Let B be an admissible A-module bialgebra. Assume that $S_H$ is an anti-bialgebra homomorphism such that ${\epsilon }_A{S}_A={\epsilon }_A$. Then we have the following equivalent items:

(1) There exists an MLHA $B⊡A$ with tensor product coproduct and counit, and

$$S(x⊡a)=\sum (S\left(a^{″}\right)\succ S\left(x\right))⊡S\left(a^{\prime }\right)$$

(6)

for any $a\in A$ and $x\in B$.

(2) For $A\in A$ and $x\in B$, we have

$$\sum a^{″}\otimes (a^{\prime }\succ x)=\sum a^{\prime }\otimes (a^{″}\succ x).$$

(7)

Proof. 

Firstly, we can define the following maps $G^{lr},G^{rl}$ from $B⊡A\otimes B⊡A$ to $B⊡A\otimes B⊡A$ by

$$\begin{array}{ccc}& & G^{lr}((x⊡a)\otimes (y⊡b))=\sum (x^{\prime }⊡a^{\prime })\otimes (a^{″}⊡a^{″})(y⊡b)\hfill \\ & & G^{rl}((x⊡a)\otimes (y⊡b))=\sum (x⊡a)(y^{\prime }⊡b^{\prime })\otimes (y^{″}⊡b^{″})\hfill \end{array}$$

for $a,b\in A$ and $x,y\in B$.

For the right counit property, we have that ${\epsilon }_{B⊡A}(x⊡a)={\epsilon }_A\left(a\right){\epsilon }_B\left(x\right)$ with $a\in A$ and $x\in B$. One needs to check ${\epsilon }_{B⊡A}{G}^{lr}((x⊡a)\otimes (y⊡b))=(x⊡a)(y⊡b)$ for any $a,b\in A$ and $x,y\in B$. It is obvious.

Denote $\tilde{\Delta }:={\Delta }_{B⊡A}$. As for the coproduct property, we have that the map $\tilde{\Delta }:B⊡A⟶M((B⊡A)\otimes (B⊡A))$ is defined via $\tilde{\Delta }(x⊡a)=\sum (x^{\prime }⊡a^{\prime })\otimes (x^{″}⊡a^{″})$ with $a\in A$ and $x\in B$. Then ${\Delta }_{B⊡A}$ is an algebra map.

Actually,

$$\begin{array}{ccc}& & \tilde{\Delta }\left((x⊡a)(y⊡b)\right)=\sum \Delta (x(a^{\prime }\succ y)⊡a^{″}b)\hfill \\ & & =\sum {(x(a^{\prime }\succ y)⊡a^{″}b)}^{\prime }\otimes {(x(a^{\prime }\succ y)⊡a^{″}b)}^{″}\hfill \\ & & =\sum x^{\prime }{(a^{\prime }\succ y)}^{\prime }⊡a^{″}{b}^{\prime }\otimes x^{″}{(a^{\prime }\succ y)}^{″}⊡a^{‴}{b}^{″}\hfill \\ & & \stackrel{\left(3\right)}{=}\sum x^{\prime }(a^{\prime }\succ y^{\prime })⊡a^{‴}{b}^{\prime }\otimes x^{″}(a^{″}\succ y^{″})⊡a^{⁗}{b}^{″}......................\left(X\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}& & \tilde{\Delta }(x⊡a)\tilde{\Delta }(y⊡b))=\sum [(x^{\prime }⊡a^{\prime })\otimes (x^{″}⊡a^{″})][(y^{\prime }⊡b^{\prime })\otimes (y^{″}⊡b^{″})]\hfill \\ & & =\sum (x^{\prime }⊡a^{\prime })(y^{\prime }⊡b^{\prime })\otimes (x^{″}⊡a^{″})(y^{″}⊡b^{″})\hfill \\ & & =\sum x^{\prime }(a^{\prime }\succ y^{\prime })⊡a^{″}{b}^{\prime }\otimes x^{″}(a^{‴}\succ y^{″})⊡a^{⁗}{b}^{″}...............\left(Y\right)\hfill \end{array}$$

for any $a,b\in A$ and $x,y\in B$.

If the formula (X) = (Y), then we have

$$\begin{array}{ccc}& & \sum x^{\prime }(a^{\prime }\succ y^{\prime })⊡a^{‴}{b}^{\prime }\otimes x^{″}(h^{″}\succ y^{″})⊡a^{⁗}{b}^{″}\hfill \\ & =& \sum x^{\prime }(a^{\prime }\succ y^{\prime })⊡a^{″}{b}^{\prime }\otimes x^{″}(a^{‴}\succ y^{″})⊡a^{⁗}{b}^{″}.\hfill \end{array}$$

Applying ${\epsilon }_B\otimes i{d}^2\otimes {\epsilon }_A$ and Equation (4) to the above equation, one has

$$\begin{array}{c}\hfill \sum a^{″}b\otimes a(a^{\prime }\succ y)=\sum a^{\prime }b\otimes a(a^{″}\succ y),\end{array}$$

which is nothing less than Equation (7) by the nondegeneracy of the product of $B\otimes A$. Conversely, if Equation (7) holds, then we have (X) = (Y), which means that $\tilde{\Delta }$ is multiplicative.

In order to show that the map $G^{lr}$ is injective and $G^{rl}$ is surjective, one defines two maps $R^{lr},R^{rl}$ in $End(B⊡A\otimes B⊡A)$ as follows:

$$\begin{array}{ccc}& & R^{lr}((a⊡h)\otimes (b⊡l))=\sum (a^{\prime }⊡h^{\prime })\otimes (S(a^{″}⊡h^{″})(b⊡l)\hfill \\ & & R^{rl}((a⊡h)\otimes (b⊡l))=\sum (a⊡h)S(b^{\prime }⊡l^{\prime })\otimes (b^{″}⊡l^{″})\hfill \end{array}$$

for any $a,b\in B$ and $h,l\in A$, and we will check that $R^{lr}{G}^{lr}=i{d}_{B⊡A\otimes B⊡A}$ and $G^{rl}{R}^{rl}=i{d}_{B⊡A\otimes B⊡A}$. In fact,

$$\begin{array}{ccc}& & \left(R^{lr}{G}^{lr}\right)((a⊡h)\otimes (b⊡l))=R^{lr}[\sum (a^{\prime }⊡h^{\prime })\otimes (a^{″}⊡h^{″})(b⊡l)]\hfill \\ & & =\sum (a^{\prime }⊡h^{\prime })\otimes (S(a^{″}⊡h^{″})(a^{″}⊡h^{‴})(b⊡l)\hfill \\ & & =\sum (a^{\prime }⊡h^{\prime })\otimes \left(S(a^{″}⊡h^{″})(a_{\left(3\right)}⊡h^{‴})\right)(b⊡l)\hfill \\ & & \stackrel{\left(6\right)}{=}\sum (a^{\prime }⊡h^{\prime })\otimes [(S\left(h^{‴}\right)\succ S\left(a^{″}\right))⊡S\left(h^{″}\right)](a_{\left(3\right)}⊡h^{‴})(b⊡l)\hfill \\ & & =\sum (a^{\prime }⊡h^{\prime })\otimes [(S\left(h^{‴}\right)\succ \left(S\left(a^{″}\right)a^{‴}\right))⊡S\left(h^{″}\right)h^{⁗}](b⊡l)\hfill \\ & & \stackrel{\left(7\right)}{=}\sum (a^{\prime }⊡h^{\prime })\otimes [(S\left(h^{″}\right)\succ \left(S\left(a^{″}\right)a^{‴}\right))⊡S\left(h^{‴}\right)h^{⁗}](b⊡l)\hfill \\ & & =\sum (a⊡h^{\prime })\otimes {\epsilon }_A\left(h^{″}\right)(b⊡l)=\sum (a⊡h)\otimes (b⊡l),\hfill \end{array}$$

and, we also compute as follows:

$$\begin{array}{ccc}& & G^{rl}{R}^{rl}((a⊡h)\otimes (b⊡l))=G^{rl}[\sum (a⊡h)S(b^{\prime }⊡l^{\prime })\otimes (b^{″}⊡l^{″})]\hfill \\ & & =\sum (a⊡h)S(b^{\prime }⊡l^{\prime })(b^{″}⊡l^{″})\otimes (b^{‴}⊡l^{‴})\hfill \\ & & =\sum (a⊡h)((S\left(l^{″}\right)\succ S\left(b^{\prime }\right))⊡S\left(l^{\prime }\right))(b^{″}⊡l^{‴})\otimes (b^{‴}⊡l^{⁗})\hfill \\ & & =\sum (a⊡h)(S\left(l^{″}\right)\succ \left(S\left(b^{\prime }\right)b^{″}\right)⊡S\left(l^{\prime }\right)l^{‴})\otimes (b^{‴}⊡l^{⁗})\hfill \\ & & \stackrel{\left(7\right)}{=}\sum (a⊡h)(S\left(l^{\prime }\right)\succ \left(S\left(b^{\prime }\right)b^{″}\right)⊡S\left(l^{″}\right)l^{‴})\otimes (b^{‴}⊡l^{⁗})=(a⊡h)\otimes (b⊡l).\hfill \end{array}$$

Actually, we have

$$\begin{array}{ccc}& & m_{B⊡A}\left((S_{B⊡A}\otimes i{d}^2)G^{lr}((a⊡h)\otimes (b⊡l))\right)\hfill \\ & & =m_{B⊡A}((S_{B⊡A}\otimes i{d}^2)[\sum (a^{\prime }⊡h^{\prime })\otimes (a^{″}(h^{″}\succ b)⊡h^{‴}l)]\hfill \\ & & \stackrel{\left(6\right)}{=}{m}_{B⊡A}(\sum (S\left(h^{″}\right)\succ S\left(a^{\prime }\right))⊡S\left(h^{\prime }\right)\otimes (a^{″}(h^{‴}\succ b)⊡h^{⁗}l))\hfill \\ & & =\sum (S\left(h^{‴}\right)\succ S\left(a^{\prime }\right))(S\left(h^{″}\right)\succ \left(a^{″}(h^{⁗}\succ b)\right))⊡S\left(h^{\prime }\right)h^{\prime \prime \prime \prime \prime }l\left)\right)\hfill \\ & & =\sum S\left(h^{″}\right)\succ \left(S\left(a^{\prime }\right)a^{″}\right)(h^{‴}\succ b)⊡S\left(h^{\prime }\right)h^{⁗}l\left)\right)\hfill \\ & & =\epsilon \left(a\right)\sum S(h^{″}{h}^{‴}\succ b)⊡S\left(h^{\prime }\right)h^{⁗}l\left)\right)={\epsilon }_{B⊡A}(a⊡h)(b⊡l),\hfill \end{array}$$

and so $S_{B⊡A}$ is a left antipode of $B⊡A$.

This completes the proof. □

Remark here that Equation (7) is similar to the condition of co-commutativity in a bialgebra.

For multiplier left Hopf algebras, there is a natural notion of left and right invariance for linear functionals (called integrals in Hopf algebra theory).

Definition 3.

Let $(A,\Delta )$ be a multiplier left Hopf algebra. Take $a\in A$ and let ψ be any linear functional on A. Then one can define the element $(\psi \otimes id)\Delta \left(a\right)$ in $M\left(A\right)$ as follows:

$$\begin{array}{ccc}& & ((\psi \otimes id)\Delta \left(a\right))b=(\psi \otimes id)G^{lr}(a\otimes b)\hfill \\ & & b((\psi \otimes id)\Delta \left(a\right))=(\psi \otimes id)G^{rr}(a\otimes b)\hfill \end{array}$$

for any $b\in A$. We say that ψ is a right invariant if $(\psi \otimes id)\Delta \left(a\right)=\psi \left(a\right)1_{M\left(A\right)}$ for all $a\in A$. A non-zero right invariant is said to be a right integral.

Similarly, for a linear functional ϕ on A, and $a\in A$, one can define $(id\otimes \varphi )\Delta \left(a\right)$. Then ϕ is said to be a left invariant if $(id\otimes \varphi )\Delta \left(a\right)=\varphi \left(a\right)1_{M\left(A\right)}$ for $a\in A$. A non-zero left invariant is called a left integral.

We now suppose that A and H are multiplier left Hopf algebras as in Theorem 1. Furthermore, we suppose that A and H have invariant functionals. In the proposition below we inquire if these integrals compose to integrals on $A⊡H$.

Proposition 2.

Take A and B as in Theorem 1. Let ${\psi }_A$ (resp. ${\psi }_B$) be a right integral of A (resp. B). Then ${\psi }_B\otimes {\psi }_A$ is a right integral of $B⊡A$. The same statement is true for the left integrals.

Proof. 

For any $a,b\in B$ and $h,l\in A$, we have that

$$\begin{array}{ccc}& & ({\psi }_B\otimes {\psi }_A\otimes i{d}^2)G^{lr}((a⊡h)\otimes (b⊡l))\hfill \\ & =& ({\psi }_B\otimes {\psi }_A\otimes i{d}^2)(\sum (a^{\prime }⊡h^{\prime })\otimes (a^{″}⊡h^{″})(b⊡l))\hfill \\ & =& ({\psi }_B\otimes {\psi }_A\otimes i{d}^2)(\sum (a^{\prime }⊡h^{\prime })\otimes (a^{″}(h^{″}\succ b)⊡h^{‴}l))\hfill \\ & =& \sum {\psi }_B\left(a^{\prime }\right){\psi }_A\left(h^{\prime }\right)(a^{″}(h^{″}\succ b)⊡h_{\left(3\right)}l)\hfill \\ & =& \sum {\psi }_B\left(a\right)b⊡{\psi }_A\left(h^{\prime }\right)h^{″}l=\sum {\psi }_B\left(a\right){\psi }_A\left(h\right)(b⊡l),\hfill \end{array}$$

and so $({\psi }_B\otimes {\psi }_A\otimes id\otimes id)\Delta (a⊡h)=({\psi }_B\otimes {\psi }_A)(a⊡h)1_{M(B⊡A)}$ as a left multiplier. As the product on $B⊡A$ is non-degenerate, they also equal as multipliers.

The proof for the left integrals is similar.

This completes the proof. □

Corollary 1.

Let A and B be left Hopf algebras with respective antipodes $S_A$ and $S_B$. Let B be an admissible A-module bialgebra. Assume $S_A$ is a bialgebra antiendomorphism of A such that ${\epsilon }_A{S}_A={\epsilon }_A$.

(1) We have the following equivalent statements:

(i) There is a left Hopf algebra $B⊡A$ with tensor product coproduct and counit, and

$$S(x⊡a)=\sum (S\left(a^{″}\right)\succ S\left(x\right))⊡S\left(a^{\prime }\right)$$

for any $a\in A$ and $x\in B$.

(ii) For $a\in A$ and $x\in B$, one has the following formula:

$$\sum a^{″}\otimes (a^{\prime }\succ x)=\sum a^{\prime }\otimes (a^{″}\succ x).$$

(2) Let ${\psi }_A$ (resp. ${\psi }_B$) be a right integral of A (resp. B). Then ${\psi }_B\otimes {\psi }_A$ is a right integral of $B⊡A$. The same statement yields for the left integrals.

4. Smash Products for Left Hopf Algebras

In this section, A always means a f.d. left Hopf algebra (f.d. LHA). Then $A^{*}$ is also f.d. LHA. By Example 1 (2), one has that A is a left $A^{*}$-module algebra under the action defined via the formula: $a^{*}\succ a=\sum \mathsf{\Omega }(a^{*},a^{″})a^{\prime }$ with $a\in A$ and $a^{*}\in A^{*}$.

Similarly, the right action of $a^{*}$ on a (denoted by $a\prec a^{*}$) is defined by

$$a\prec a^{*}=\sum \mathsf{\Omega }(a^{*},a^{\prime })a^{″}.$$

Lemma 1.

Let A be a f.d. LHA. For any $a^{*},b^{*}\in A^{*}$ and $a\in A$, we have

(1) $a^{*}\succ (b^{*}\succ a)=\left(a^{*}{b}^{*}\right)\succ a$;

(2) $(a\prec a^{*})\prec b^{*}=a\prec \left(a^{*}{b}^{*}\right)$.

Proof. 

(1) For item (1), we do a calculation as follows:

$$\begin{array}{ccc}\hfill a^{*}& \succ & (b^{*}\succ a)=\sum \langle b^{*},a^{″}\rangle \mathsf{\Omega }(a^{*},{\left(a^{\prime }\right)}^{″}\rangle {\left(a^{\prime }\right)}^{\prime }\hfill \\ & =& \sum \langle b^{*},a^{‴}\rangle \mathsf{\Omega }(a^{*},a^{″}\rangle a^{\prime }=\sum \langle a^{*}{b}^{*},a^{″}\rangle a^{\prime }=\left(a^{*}{b}^{*}\right)\succ a.\hfill \end{array}$$

(2) This item will be checked by a straight computation. for any $a^{*},b^{*}\in A^{*}$ and $a\in A$. □

Lemma 2.

Let A be a f.d. LHA. For any $a^{*}\in A$ and $a,b\in A$ we have

(1) $\Delta (a^{*}\succ a)=\sum a^{\prime }\otimes (a^{*}\succ a^{″})$.

(2) $\Delta (a\prec a^{*})=\sum (a^{\prime }\prec a^{*})\otimes a^{″}$.

(3) $a^{*}\succ \left(ab\right)=\sum ({\left(a^{*}\right)}^{\prime }\succ a)({\left(a^{*}\right)}^{″}\succ b)$.

(4) $\left(ab\right)\prec a^{*}=\sum (a\prec {\left(a^{*}\right)}^{\prime })(b\prec {\left(a^{*}\right)}^{″})$.

Proof. 

(1) For (1), we do a calculation as follows:

$$\begin{array}{c}\hfill \Delta (a^{*}\succ a)=\sum \langle a^{*},a^{‴}\rangle a^{\prime }\otimes a^{″}=\sum a^{\prime }\otimes (a^{*}\succ a^{″}).\end{array}$$

(2), (3) and (4) follow similarly. □

Let A be an LHA. If B is a left A-module algebra, then a smash product $B⊡A=B\otimes A$ with a multiplication described by

$$\begin{array}{c}\hfill (x⊡a)(y⊡b)=\sum x(a^{\prime }\succ y)⊡a^{″}b\end{array}$$

(8)

for any $a,b\in A$ and $x,y\in B$.

The following example is similar to ([10], Example 2(2)).

Example 2.

Let A be a f.d. LHA. Then the A can be made into a left $A^{*}$-module algebra with ≻. So, the multiplication in $H⊡H^{*}$ is given by:

$$(a⊡a^{*})(b⊡b^{*})=\sum a({\left(a^{*}\right)}^{\prime }\succ b)⊡{\left(a^{*}\right)}^{″}{b}^{*}$$

(9)

for any $a,b\in A$ and $a^{*},b^{*}\in A^{*}$.

Proposition 3.

The $(B⊡A,1_B⊡1_A)$ is a unital associative algebra if and only if A is a bialgebra and B is a left A-module algebra.

Proof. 

Firstly, it is straightforward to check that $1_B⊡1_A$ is an identity of $B⊡A$. For associativity, one has

$$\begin{array}{ccc}& & \left[(x⊡a)(y⊡b)\right](z⊡c)=\sum [x(a^{\prime }\succ y)⊡a^{″}b](z⊡c)\hfill \\ & =& \sum \left[x(a^{\prime }\succ y)\right][{\left(a^{″}b\right)}^{\prime }\succ z]⊡{\left(a^{″}b\right)}^{″}c\hfill \\ & =& \sum \left[x(a^{\prime }\succ y)\right][a^{″}{b}^{\prime }\succ z]⊡\left[a^{‴}{b}^{″}\right]c\hfill \end{array}$$

and similarly, we have

$$(x⊡a)\left[(y⊡b)(z⊡c)\right]=\sum x[(a^{\prime }\succ y)(a^{″}\succ b^{\prime }(\succ z)]⊡a^{‴}\left(b^{″}c\right)$$

for any $a,b,c\in A$ and $x,y,z\in B$. Then the associativity in A and B, and the module associativity, implies that the above two equations are equal.

Finally, if $B⊡A$ is associative then one gets:

$$\sum \left[x(a^{\prime }\succ y)\right][a^{″}{b}^{\prime }\succ z]⊡\left[a^{‴}{b}^{″}\right]c=\sum x\left[(a^{\prime }\succ y)(a^{″}\succ (b^{\prime }\succ z))\right]⊡a^{‴}\left[b^{″}c\right].$$

By putting $x=y=1$ in the above equation, we obtain:

$$\left(a^{\prime }{b}^{\prime }\right)\succ z\otimes \left[a^{″}{b}^{″}\right]c=(a^{\prime }\succ (b^{\prime }\succ z))\otimes a^{″}\left[b^{″}c\right].$$

By taking $z=1$ and applying $(id\otimes \epsilon )$ to the above equation, we have that B is a left A-module. Similarly, we can get that A is a bialgebra and B is a left A-module algebra. On the contrary, the same applies. □

Proposition 4.

Let A be a f.d. LHA. If B is a left A-module algebra, then $B⊡A$ has a left $A^{*}$-module algebra structure as follows:

$$a^{*}·(x⊡a)=x⊡(a^{*}\succ a)$$

for any $a^{*}\in A^{*},x\in B$ and $a\in A$.

Proof. 

By Lemma 1, we have that $A⊡H$ is a left $H^{*}$-module.

As for (2), for any $a^{*}\in A^{*}$, we have

$$a^{*}·(1_B⊡1_A)=1_B⊡(a^{*}\succ 1_A)=\langle a^{*},1_A\rangle 1_B⊡1_A.$$

As for Equation (1), for $a^{*}\in A^{*},x,y\in B$ and $a,b\in A$,

$$\begin{array}{ccc}& & a^{*}·\left[(x⊡a)(y⊡b)\right]=\sum a^{*}·[x(a^{\prime }·y)⊡a^{″}b]\hfill \\ & =& \sum x(a^{\prime }·y)⊡a^{*}\succ \left(a^{″}b\right)=\sum x(a^{\prime }·y)⊡({\left(a^{*}\right)}^{\prime }\succ a^{″})({\left(a^{*}\right)}^{″}\succ b)\hfill \\ & =& \sum x(a^{\prime }·y)⊡a^{″}{b}^{\prime }\mathsf{\Omega }({\left(a^{*}\right)}^{\prime },a^{‴})\mathsf{\Omega }({\left(a^{*}\right)}^{″},b^{″})\hfill \\ & =& \sum (x⊡a^{\prime })(y⊡b^{\prime })\mathsf{\Omega }({\left(a^{*}\right)}^{\prime },a^{″})\mathsf{\Omega }({\left(a^{*}\right)}^{″},b^{″})\hfill \\ & =& \sum [{\left(a^{*}\right)}^{\prime }·(x⊡a)][{\left(a^{*}\right)}^{″}·(y⊡b)]\hfill \end{array}$$

and this proves the requested result. □

5. BM Duality

This section mainly studies our BM duality theorem for finite-dimensional left Hopf algebra A of dimension $n<\infty$. In this case, let ${\{a_i^{*},a_j\}}_{i,j=1,2,\dots ,n}$ be the dual basis of $A^{*}$ and A, i.e., we have $\mathsf{\Omega }(h_i^{*},h_j)={\delta }_{ij}$ for $1\le i,j,\le n$.

Let B be a left A-module algebra with a module structure map ·. Define

$${\mho }^{BA}:B⊡A⟶End\left(B\right)$$

(10)

by ${\mho }^{BA}(x⊡a)\left(y\right)=x(a·y)$ for any $a\in A$ and $x,y\in B$.

Lemma 3.

The above map ${\mho }^{BA}$ is multiplicative.

Proof. 

Obviously, ${\mho }^{BA}(1_B⊡1_A)\left(1_B\right)=1_B$ holds. By the associativity of B, one has

$$\begin{array}{ccc}& & {\mho }^{BA}\left[(x⊡a)(y⊡b)\right]\left(z\right)=\sum {\mho }^{BA}(x(a^{\prime }·y)⊡a^{″}b)\left(z\right)\hfill \\ & =& \sum \left[(x\left(a^{\prime }\right)·y)\right][\left(a^{″}b\right)·z]=\sum x\left[(a^{\prime }·y)(a^{″}·(b·z))\right]\hfill \\ & =& x[a·\left[y(b·z)\right]]=[{\mho }^{BA}(x⊡a)\circ {\mho }^{BA}(y⊡b)]\left(z\right),\hfill \end{array}$$

for $a,b\in A$ and $x,y,z\in B$.

This ends the proof. □

Remark 3.

Let A be a f.d.LHA. By Example 2, $A⊡A^{*}$ is a smash product. Define

$${\Re }^{A{A}^{*}}:A^{*}⟶End\left(A\right)$$

(11)

via ${\Re }^{A{A}^{*}}\left(a^{*}\right)\left(a\right)=a\prec a^{*}$ for $a\in A$ and $a^{*}\in A^{*}$. Then ${\Re }^{A{A}^{*}}$ maintains the identity element property and is anti-multiplicative.

Given a left A-module algebra B, we get: $B⊡A$. By using the map ℧ from (10) and Proposition 4, we can define:

$$F={\mho }^{B⊡A}:(B⊡A)⊡A^{*}⟶End(B⊡A).$$

and

$$P={\lambda }_l\otimes {\mho }^{A{A}^{*}}:B\otimes (A⊡A^{*})⟶End(B⊡A)$$

where ${\lambda }_l:B\hookrightarrow End\left(B\right)$ is given by $x\mapsto (a\mapsto xa)$.

We remark here that $id\otimes {\Re }^{A{A}^{*}}\left(a^{*}\right)=P(1\otimes u)$ for some $u\in A⊡A^{*}$ for $a^{*}\in A^{*}$.

The following result is different from the case of Hopf quasigroups. In that case, we know that F and P do not preserver multiplication ([7], Lemma 3.2).

Lemma 4.

(1) F and P are multiplicative.

(2) We have the following identities: for $u=(x⊡a)⊡a^{*}$ and $v=x⊡(a⊡a^{*})$ with $x\in B,a\in A,a^{*}\in A^{*}$,

$$\begin{array}{ccc}& \left(a\right)& F\left(u\right)=F((x⊡1)⊡\epsilon )\circ F((1⊡a)⊡a^{*}),\hfill \\ & \left(b\right)& P\left(v\right)=P\left(xotimes(1⊡\epsilon )\right)\circ P(1\otimes (a⊡a^{*})).\hfill \end{array}$$

Proof. 

(1) We only show that F is a multiplicative, similar as for P. Actually, for $w=y⊡(b⊡b^{*})$ with $y\in B,b\in A,b^{*}\in A^{*}$, one has

$$\begin{array}{ccc}& & F\left[uw\right](z⊡c)\hfill \\ & =& \sum (x(a^{\prime }\succ y)⊡a^{″}{b}^{\prime })(z⊡{\left(a^{*}\right)}^{″}{b}^{*}\succ c)\mathsf{\Omega }({\left(a^{*}\right)}^{\prime },b^{″})\hfill \\ & =& \sum F\left[((x⊡a)⊡a^{*})\right](y(b^{\prime }\succ z)⊡b^{″}(b^{*}\succ c))\hfill \\ & =& F((x⊡a)⊡a^{*})F((y⊡b)⊡b^{*})(z⊡c).\hfill \end{array}$$

(2) As for (a), we have

$$\begin{array}{ccc}& & F\left(u\right)(z⊡c)=(x⊡a)(z⊡(a^{*}\succ c))\hfill \\ & =& x(h^{\prime }\succ z)⊡a^{″}(a^{*}\succ c))=(x⊡c)[(a^{\prime }\succ z)⊡a^{″}(a^{*}\succ c)]\hfill \\ & =& \left[F((a⊡1)⊡\epsilon )\right((1⊡a)(z⊡a^{*}\succ c)=[F((a⊡1)⊡\epsilon )\circ F((1⊡a)⊡a^{*})](z⊡c).\hfill \end{array}$$

Similarly for (b),

$$\begin{array}{ccc}& & F\left(u\right)(z⊡c)=xz⊡a(a^{*}\succ c)\hfill \\ & =& [P(x\otimes (1⊡\epsilon ))(z\otimes a(a^{*}\succ c))[P(x\otimes (1⊡\epsilon ))\circ P(1\otimes (a⊡a^{*}))](z⊡c).\hfill \end{array}$$

□

In what follows, let A be a f.d. LHA with a bijective left antipode S such that S is a bialgebra antihomomorphis.

Lemma 5.

F and P are injective linear maps.

Proof. 

Firstly, for all $x,y\in B,a,b\in A,a^{*}\in A^{*}$ and $f\in End(B⊡A)$, one defines

$$F^{\prime }:(B⊡A)⊡A^{*}⟶End(B⊡A),F^{\prime }\left(u\right)(z⊡c)=\mathsf{\Omega }(a^{*},c)(x⊡a)(z⊡c).$$

For an element $x={\sum }_{i=1}^n{y}_i⊡h_i^{*}\in Ker\left(F^{\prime }\right)$, here $y_i\in A⊡H$, and ${\left\{h_i^{*}\right\}}_{i=1,2,\dots ,n}$ is a linearly independent subset of $A^{*}$. Choose $a_1 ,a_2 ,·a_n$ such that $a_i^{*}\left(a_j\right)={\delta }_{ij}$, with $1\le i,j,\le n$. Then $0=F^{\prime }\left(x\right)(1⊡k_i)=y_i$ for all i, thus $x=0$. This proves that $F^{\prime }$ is injective. Similarly, we define

$$P^{\prime }:B\otimes (A⊡A^{*})⟶End(B⊡A),P^{\prime }\left(u\right)(z⊡c)=\mathsf{\Omega }(a^{*},c)(xz⊡a)$$

and we can check that $P^{\prime }$ is injective. Define

$$\begin{array}{ccc}& & W:End(B⊡A)⟶End(B⊡A),W\left(f\right)(z⊡c)=\sum \left[f(z⊡c^{″})\right](1⊡c^{\prime })\hfill \end{array}$$

In order to show that W is injective, one defines $U:End(B⊡A)⟶End(B⊡A)$ by

$$U\left(f\right)(z⊡c)=\sum \left[f(z⊡c^{″})\right](1⊡S^{-1}\left(c^{\prime }\right)).$$

Then

$$\begin{array}{ccc}& & (U\circ W)\left(f\right)(z⊡c)=\sum \left[W\left(f\right)(b⊡c^{″})\right](1⊡S^{-1}\left(c^{\prime }\right))\hfill \\ & =& \sum \left[\left[f(z⊡c^{‴})\right](1⊡c^{″})\right](1⊡S^{-1}\left(c^{\prime }\right))\hfill \\ & =& \sum u_i⊡\left(v_i{c}^{″}{S}^{-1}\left(c^{\prime }\right)\right)=\sum u_i⊡v_i\epsilon \left(l\right)=\sum f(z⊡c^{″})\epsilon \left(c^{\prime }\right)=f(z⊡c).\hfill \end{array}$$

By analogy, one gets: $(W\circ U)\left(f\right)(z⊡c)=f(b⊡l)$. Thus, U is a two-sided inverse for W.

For $F=W\circ F^{\prime }$ and $P=W\circ P^{\prime }$, we only show the first one; the second one can be obtained similarly. In fact,

$$\begin{array}{ccc}& & (W\circ F^{\prime })\left(u\right)(z⊡c)=W\left[F^{\prime }((x⊡a)⊡a^{*})\right](z⊡c)\hfill \\ & =& \sum F^{\prime }\left(u\right)(z⊡c^{″})(1⊡c^{\prime })=\sum \left[(x⊡a)(z⊡1)\right](1⊡c^{\prime })\mathsf{\Omega }(a^{*},c^{″})\hfill \\ & =& (x⊡a)(z⊡(a^{*}\succ c))=F\left(u\right)(z⊡c),\hfill \end{array}$$

and so F and P are injective. □

Corollary 2.

The map ${\mho }^{A{A}^{*}}$ is bijective.

Proof. 

By using Example 2, one has $\mathbf{k}⊡A\cong A$ and $P={\mho }^{A{A}^{*}}$, such that ${\mho }^{A{A}^{*}}$ is injective. We also have that $dim(A⊡A^{*})=n^2=dim\left(End\left(A\right)\right)$. Therefore, ${\mho }^{A{A}^{*}}$ is a bijective linear map. Thus, one gets:

$$A⊡A^{*}\cong End\left(a\right)\cong M_n\left(\mathbf{k}\right)$$

The proof is complete. □

One remarks that Corollary 2 was shown for finite-dimensional Hopf algebras by M. Van den Bergh in [19].

We define a map $Q\in End(B⊡A)$ by

$$Q(z⊡c)=\sum (S^{-1}\left(c^{\prime }\right)\succ z)⊡c^{″}$$

for $c\in A$ and $z\in B$.

Lemma 6.

The map Q is bijective.

Proof. 

In order to show that Q is invertible, we define O by $O(z⊡c)=\sum (c^{\prime }\succ z)⊡c^{″}$ for any $z\in B$ and $c\in A$. A computation is given as follows:

$$\begin{array}{ccc}& & Q\left(O(z⊡c)\right)=\sum Q[(c^{\prime }\succ z)⊡c^{″}]\hfill \\ & =& \sum (S^{-1}\left(c^{″}\right)\succ (c^{\prime }\succ z))⊡c^{‴}=\sum \epsilon \left(c^{\prime }\right)z⊡c^{″}=\sum z⊡c,\hfill \end{array}$$

thus, $Q\circ O=id$. Similarly, $O\left(Q\right(z⊡c\left)\right)=\sum z⊡c$, thus, $O\circ Q=id$. The proof is complete. □

Remark 4.

Let $x\in B$ and $x_i=a_i\succ x$ with $i\in \{1,2,\dots ,n\}$. For $a\in A$, set $a={\sum }_i{p}_i{a}_i$ with $p_i\in \mathbb{F}$. Then $a\succ x={\sum }_i{p}_i(a_i\succ x)={\sum }_i{p}_i{x}_i$. But $\mathsf{\Omega }(a_j^{*},a)=\mathsf{\Omega }(a_j^{*},{\sum }_i{p}_i{a}_i)=p_j$, and so

$$a\succ x=\sum _i\mathsf{\Omega }(x_i^{*},a)x_i.$$

(12)

Let $x\in B$ so that $x_i=a_i\succ x$ with $i\in \{1,2,\dots ,n\}$ and for any $b\in A,h,l\in H$ and $h^{*}\in H^{*}$. Then we have lemma.

Lemma 7.

The following equations hold:

$$\begin{array}{ccc}& \left(a\right)& O\circ P(1\otimes (1⊡a^{*}))\circ Q=F((1⊡1)⊡a^{*}),\hfill \\ & \left(b\right)& O\circ P(x\otimes (1⊡\epsilon ))\circ Q=\sum _{i=1}^{n}F((x_i⊡1)⊡\epsilon )\circ O\circ (id\otimes {\Re }^{A{A}^{*}}\left(a_i^{*}\right))\circ Q,\hfill \\ & \left(c\right)& Q\circ F((x⊡1)⊡\epsilon )\circ O=\sum _{i=1}^{n}P((x_i\otimes (1⊡\epsilon ))\circ (id\otimes {\Re }^{A{A}^{*}}\left(S^{*-1}\left(a_i^{*}\right)\right),\hfill \\ & \left(d\right)& [O\circ P(1\otimes (a⊡\epsilon ))\circ Q](z⊡c)=F((1⊡a{c}^{″})⊡\epsilon )(S^{-1}(c^{\prime }\succ z)⊡1).\hfill \end{array}$$

Proof. 

For (a), we have, for any $z\in B$ and $c\in A$,

$$\begin{array}{ccc}& & [O\circ P(1\otimes (1⊡a^{*}))\circ Q](z⊡c)=\sum [O\circ P(1\otimes (1⊡a^{*}))](S^{-1}\left(c^{\prime }\right)\succ z)⊡c^{″}\hfill \\ & =& \sum O[(S^{-1}\left(c^{\prime }\right)\succ z)⊡(a^{*}\succ c^{″})]\hfill \\ & =& \sum [{\left[(x^{*}\succ c^{″})\right]}^{\prime }\succ (S^{-1}\left(c^{\prime }\right)\succ z)]⊡{\left[(a^{*}\succ c^{″})\right]}^{″}\hfill \\ & =& \sum [{(a^{*}\succ c^{″})}^{\prime }\succ (S^{-1}\left(c^{\prime }\right)\succ z)]⊡\left[{(a^{*}\succ c^{″})}^{″}\right]\hfill \\ & =& \sum [c^{‴}\succ \left(S^{-1}(c^{\prime }\succ z)\right]⊡(a^{*}\succ c^{″})\hfill \\ & =& \sum (z⊡(a^{*}\succ c)=F((1⊡1)⊡a^{*})(z⊡c).\hfill \end{array}$$

Similarly for (b). As for (c), one computes:

$$\begin{array}{ccc}& & [Q\circ F((x⊡1)⊡\epsilon )\circ O](z⊡c)=\sum [Q\circ F((x⊡1)⊡\epsilon )]((c^{\prime }\succ b)⊡c^{″})\hfill \\ & =& \sum Q\left[(x⊡1)((c^{\prime }\succ b)⊡c^{″})\right]=\sum Q[x(c^{\prime }\succ z)⊡c^{″}]\hfill \\ & =& \sum S^{-1}\left(c^{″}\right)\succ \left[x(c^{\prime }\succ z)\right]⊡c^{‴}]\hfill \\ & =& \sum _{i=1}^n\left[P\right((x_i\otimes (1⊡\epsilon ))\circ (id\otimes {\Re }^{A{A}^{*}}\left(S^{*-1}\left(a_i^{*}\right)\right)](z⊡c).\hfill \end{array}$$

Similarly for (d). The proof is complete. □

Remark 5.

Generally, for any $a\in A$ and $a^{*}\in A^{*}$, we have $O\circ P(1\otimes (a⊡a^{*}))\circ Q\ne F((1⊡a)⊡a^{*})$.

Given a f.d. LHA A which acts on an algebra B, we will prove that Blattner–Montgomery’s theorem (see, [6]) is still true in the following main theorem.

Theorem 2.

Let A be a f.d. LHA with a bijective left antipode S such that S is a bialgebra antihomomorphis and B a left H-module algebra. Then

$$(B⊡A)⊡A^{*}\cong B\otimes (A⊡A^{*}).$$

Proof. 

For this result, we prove that $O\circ P(x\otimes (a⊡a^{*}))\circ Q$ belong to $F((B⊡A)⊡A^{*})$, for $x\in B,a\in A$ and $a^{*}\in A^{*}$.

From the equation $u=(x\otimes (1⊡\epsilon ))(1\otimes (a⊡a^{*}))$ and Lemma 5, one gets

$$\begin{array}{ccc}& & O\circ P\left(u\right)\circ Q=O\circ P(x\otimes (1⊡\epsilon ))\circ P(1\otimes (a⊡a^{*}))\circ Q\hfill \\ & =& [O\circ P(x\otimes (1⊡\epsilon ))\circ Q]\circ [O\circ P(1\otimes (a⊡a^{*}))\circ Q]\text{from Lemma 6}.\hfill \end{array}$$

So, it is straightforward to check $\mathsf{\Omega }\circ P(a\otimes (1⊡\epsilon \left)\right)\circ Q$ (by Lemma 7(b)) and $O\circ P(1\otimes (h⊡h^{*}))\circ Q$ (by Lemma 7(a)(d)) are all in $F((A⊡H)⊡H^{*})$.

Similarly, $Q\circ F((x⊡a)⊡a^{*})\circ \mathsf{\Omega }$ belongs to $P(B\otimes (A⊡A^{*}))$ by Lemma 7(a)(c)(d).

Finally, one gets

$$F((B⊡A)⊡A^{*})=Q^{-1}\circ P(B\otimes (A⊡A^{*}))\circ Q.$$

This finishes the proof. □

Corollary 3.

Let A be a f.d. LHA with a bijective left antipode S such that S is a bialgebra antihomomorphis and B a left A-module algebra. Then

$$(B⊡A)⊡A^{*}\cong B\otimes (A⊡A^{*})\cong B\otimes M_n\left(\mathbf{k}\right)\cong M_n\left(A\right).$$

Proof. 

This conclusion can be directly obtained from Theorem 2 and Corollary 3. The proof is complete. □

6. Conclusions and Further Research

In this paper, we have two main results presented in Theorems 1 and 2. The former theorem gives an approach to get a new multiplier left Hopf algebra, and the latter theorem obtains a duality theorem for a smash product of f.d. LHAs acting on an algebra. These results generalize the existing work in the references.

During our research process, we found that the following questions are worth studying.

Question 1: How do we construct the Drinfeld double for left Hopf algebras and multiplier left Hopf algebras?

In fact, if we look again at the arguments in Theorem 1, they seem to support such a conjecture.

Question 2: How do we define a quasitriangular left Hopf algebra and what Yang–Baxter equation does it correspond to?

It is well-know that the quasitriangular structure can give rise to the solution of quantum Yang-Baxter equation.

Question 3: Does the Pontryagin duality hold for multiplier left Hopf algebras?

In our work, we have studied the integrals in Proposition 3. For the case of multiplier Hopf algebras, Van Daele solves the Pontryagin duality through the tool of integrals (see [14]).